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Z 2 O 6 3- (aq) + JO 2 - (aq)  J 2 O 7 2- (aq) + Z 7 O 4 3+ (aq) 68H + (aq) + 41e - +7 Z 2 O 6 3- (aq)  2Z 7 O 4 3+ (aq) + 34H 2 O(l) 3H 2 O(l) + 2JO.

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Presentation on theme: "Z 2 O 6 3- (aq) + JO 2 - (aq)  J 2 O 7 2- (aq) + Z 7 O 4 3+ (aq) 68H + (aq) + 41e - +7 Z 2 O 6 3- (aq)  2Z 7 O 4 3+ (aq) + 34H 2 O(l) 3H 2 O(l) + 2JO."— Presentation transcript:

1 Z 2 O 6 3- (aq) + JO 2 - (aq)  J 2 O 7 2- (aq) + Z 7 O 4 3+ (aq) 68H + (aq) + 41e - +7 Z 2 O 6 3- (aq)  2Z 7 O 4 3+ (aq) + 34H 2 O(l) 3H 2 O(l) + 2JO 2 - (aq)  J 2 O 7 2- (aq) +6H + (aq) + 6e - 82JO 2 - (aq) + 42Z 2 O 6 3- (aq) + 162H + (aq)  41J 2 O 7 2- (aq)+ 12Z 7 O 4 3+ (aq) + 81H 2 O(l)

2 REDOX REACTIONS TODAY: We’re going to perform calculations to determine the quantities of substances involved in redox titrations In other words we’re going to learn the stoich involved for our lab

3 REDOX STOICHIOMETRY

4 Stoichiometry the study of the quantitative aspects of chemical reactions how much of each substance we need in a reaction

5 Stoich allows us to use concentration equations to find unknown value in REDOX

6

7 Titration Slowly add 1 solution (titrant) to another solution (sample) As the titrant is added to the sample, the 2 solutions react

8 As more titrant is added a redox reaction occurs until equilibrium is reached and the reaction can no longer proceed

9 2 strong oxidizing agents often used are solutions containing permanganate ions (MnO 4 - ) or dichromate ions (Cr 2 O 7 2- )

10 The endpoint can be seen when the addition of ONE DROP of the titrant causes a permanent color change in the sample. That is when no more of the sample is available to react with the titrant

11 Titration Calculations Using the known concentration and volume of the titrant we can find the moles used Using the stoichiometric ratio we can find the moles of the sample reacting

12 With the moles of the sample known, we can use our known volume to find the concentration of the sample

13 Very Important! The only difference in redox stoich vs regular stoich is you determine the balanced equation through half reactions and the net equation!

14 Example 1: Find the concentration of Fe2+(aq) in a 10.0mL sample that requires 14.0mL of a 0.050mol/L acidic permanganate solution to reach the endpoint.

15 Example 2: Find the concentration of Sn 2+ (aq) in a 50.0mL sample that requires 64.8mL of a 0.080mol/L acidic dichromate solution to reach the endpoint.


Download ppt "Z 2 O 6 3- (aq) + JO 2 - (aq)  J 2 O 7 2- (aq) + Z 7 O 4 3+ (aq) 68H + (aq) + 41e - +7 Z 2 O 6 3- (aq)  2Z 7 O 4 3+ (aq) + 34H 2 O(l) 3H 2 O(l) + 2JO."

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